THE LEGENDRE AND LAGUERRE POLYNOMIALS & THE

ELEMENTARY QUANTUM MECHANICAL MODEL OF THE

HYDROGEN ATOM

TIMOTHY JONES

Abstract. In this report, we explore the basic Quantum Mechanical analy-sis of hydrogen. In the process we come across the Legendre and Laguerrepolynomials, and present an informal derivation of these functions and theirnormalization.

1. The Bohr model

We first consider an early model of the hydrogen atom, (Thompson 1903) [1]. Inthis model, the atom is a blob of uniformly distributed positive charge, a sphere ofradius 1A and charge +e. The electron is considered to exist as a point like particlewithin this sphere. Breaking the sphere up into infinitesimal shells, each shell willcontribute a potential to the electron of

(1.1) dV =e2r2dr

40

pi0

sin d

r2 + r2 2rr cos Where = e/(4R3/3), r is the radius of the shell, and r is the location of theelectron relative to the center of the sphere. Considering shells of radius less thanthe position of the electron, and those greater, the potential is(1.2)

V =e440

( r0

r2dr

r+

Rr

r2dr

r

)=e440

(r26

+R2

2

) F = e

2

40R3r

Similar in form to the harmonic potential, F = kr, we have,

(1.3) =

k

me=

e2

40R3me

We divide by 2 and use known constants to find that we expect a frequency of2.43 1015Hz corresponding to 123 nm wavelength. This is approximately theLyman-alpha wavelength (121.5 nm), but does not account for the infinite spec-trum that Hydrogen actually exhibits. Nor does this model incorporate Plancksdiscovery of quantization of radiation.

The earliest model of the hydrogen atom that accounted for Plancks discov-ery that a proper explanation of the blackbody radiation spectrum could only beachieved through quantization of the energy released by atoms was achieved byNiels Bohr. The model had the following properties, as summarized by Bohr [2, 3]:

(1) The electron emits radiation when transitioning from one discreet state to the next.(2) Classical mechanics is valid when the electron is not transitioning.(3) In transition from a state to another, energy differences being E, a photon of frequency

= E/h is emitted.(4) Angular momentum is quantized and identifies permitted orbits. It is always a natural

number multiple of h/2pi.

The model is amazingly simple, but manages to correctly predict the ground stateenergy level. It goes as follows. Since we assume the laws of classical mechanics

1

hold in non-transition periods, the electron obeys the Coulomb force law and thecentripetal acceleration formulation:

(1.4)mev

2

r=

1

4o

e2

r2

Under the quantization of orbital angular momentum, we can write L = mevr =n~, n Z > 0. This number, n, will be seen throughout this derivation as a key andimportant variable, the angular momentum quantum number. From our definitionof L, we find that,

(1.5)mev

2

r=

mer

(n2~2

m2er2

)=

1

4o

e2

r2 r = 4on

2~2

e2me

The smallest possible orbit is when n = 1. This is called the Bohr radius, and isgiven as

(1.6) ao =4o~

2

e2me

Energy is the sum of its Kinetic and Potential partitions, and given our previousresults,

(1.7) En =mev

2e

2 e

2

4or= me

2

(e2

4o~n

)2It was a great result that this formula predicts the ground state energy and excitedenergy states of the electron:

(1.8) E1 = me2

(e2

4o~

)2= 2.18E 18J = 13.6ev

The model also gives a more accurate picture of the spectrum of hydrogen, wherewe can use En = ~ to find the corresponding frequencies. But this is not goodenough. This model tells us nothing about why and how the transitions are made.It is rather ad hoc.

Bohr also introduced a helpful principle in Quantum Mechanics, the Correspon-dence Principle, the idea that Quantum Mechanical calculations must limit to clas-sical results when the census of quantum numbers tends towards infinity.

2. The Hydrogen Calculation using Schrodingers equation

Enter Schrodinger. He proposed that an object in quantum mechanics obeys thewave equation,

(2.1)2

x2= i~

t

But of course, the Hydrogen atom does not live in one dimension. We must moveto three dimensions, using spherical coordinates as the most natural setting for ourderivation.

In spherical coordinates the Laplacian becomes:

(2.2) 2 = 1r2

r

(r2

r

)+

1

r2 sin()

(sin()

)+

1

r2 sin2()

(2

2

)Schrodingers equation becomes:

~2

2m

(1

r2

r

(r2

r

)+

1

r2 sin()

(sin()

)

+1

r2 sin2()

(2

2

))+ V = E(2.3)

2

If we assume that the solutions can be found through separation of variables:

(r, , ) = R(r)Y (, ) ~

2

2m

(1

R

d

dr

(r2dR

dr

))+

1

Y sin()

(sin()

Y

)

+1

Y sin()

2

2+ r2(V E) = 0

= 1R

(r2dR

dr

) 2mr

2

~2(V E) = ,

1

Y

(1

sin()

(sin()

Y

)+

1

sin2()

2Y

2

)= (2.4)

We have divided, and now we are ready to solve. The radial term contains thepotential V, which is a function of r for hydrogen. It is the more complex of thetwo, and so we seek to explore the angular term first. Our analysis will find the constant of separation in its exact form, enabling a strong attempt at solving theradial equation.

3. The Angular Component

Once again, it is reasonable to assume a certain dimensional independence whichenables separation of variables in terms of the and components. Under thisregime:

Y (, ) = ()() =1

(sin()

d

d

(sin()

d

d

)+sin2

)+

1

d2

d2= 0

1

(sin()

d

d

(sin()

d

d

)+sin2

)= m2,

1

d2

d2= m2 = () = eim, m Z(3.1)

The side of the equation was quickly solved. We can recast this equation in thefollowing variations, where we let C represent x = cos() and S

1 x2 = sin()

; we introduce the notation H which represents an introduction of a new idea intothe stream of derivation, and is followed by the immediate consequence of thisnew idea; we then consider the m = 0 case.

Sd

d

(Sd

d

)+ (S2) = 0

x = C() dxd

= S() = S

(Sd

dx

dx

d

)

= Sd

d

(S2d

dx

)= 2S2C d

dx+

(S3 d

dx

d

d= S4

d2

dx2

)

H = y S2y 2Cy +y= (1 x2)y 2xy +y = d

dx

((1 x2)y)+y = 0

As we further press on this equation, we shall find it suggesting its own solution,even betraying the separation constant .

3

3.1. A closer look. Let us take a careful look at the equation,

(3.2) (1 x2)y 2xy +y = 0We note that the x derivative of (1 x2) is 2x. This is highly suggestive that thekernel of our function is (1 x2) itself. The multiple derivatives suggest somethinga la w = (1 x2)n. So let us explore this kernel. Its first derivative is w =2xn(1 x2)n1 or w(1 x2) + 2nxw = 0 Define Dn to be the nth derivative ofthis latter equation. Then we have the following formulation:

D0 : (1 x2)w + 2nxw = 0D1 : (1 x2)w 2xw + 2nxw + 2nw

= (1 x2)w + 2x(n 1)w + 2nw = 0D2 : (1 x2)w 2xw + 2x(n 1)w + 2(N 1)w + 2nw

= (1 x2)w + 2x(n 2)w + (4n 2)w = 0D3 : (1 x2)w(4) 2xw(3) + 2x(n 2)w(3) + 2(n 2)w(2) + (rn 2)w(2)

= (1 x2)w(4) + 2x(n 3)w(3) + (6n 6)w(2) = 0(3.3)

If we have instead w = (x2 1)n:D0 : (x

2 1)w 2nxw = 0D1 : (x

2 1)w 2(n 1)xw 2nxw = 0D2 : (x

2 1)w(3) 2(n 2)xw(2) 2(2n 1)w = 0D3 : (x

2 1)w(4) 2(n 3)xw(3) 2(3n 3)w(2) = 0...

... =...

D(n+1) : (x2 1)w(n) + 2xw(n) 2 (n2 nm=0(m 1))wn = 0(x2 1)w(n) + 2xw(n) 2

(n2 n(n1)2

)wn = 0

(x2 1)w(n) + 2xw(n) n(n+ 1)wn = 0(3.4)

This latter equation is in the same form as our original equation. That is,

(3.5) y =d2

dxn(x2 1)n

Or, if we so choose,

(3.6) y =1

2nn!

d2

dxn(x2 1)n

This is the so called Legendre Polynomial, denoted by Pn(x) We so assign =l(l+1). But what of the m 6= 0 cases? In such a situation, our equation of interestcan be cast as:

(3.7) (1 x2)y 2xy +(l(l + 1) m

2

1 x2)y = 0

Let us continue the derivation series from above; Recall that

(3.8) D(n+1) : (x2 1)w(n) + 2xw(n) n(n+ 1)wn = 0

Continuing the process we get,

(3.9) Dk,n+1 : (1 x2)y(2+k) x(2 + 2k)y(k+1) + (l(l + 1) k(k + 1)) y(k) = 04

Or, in the spirit of what we have done thus far, write:

(3.10) z =

(dk

dxk

)y (1 x2)z 2x(k + 1)z + (l(l + 1) k(k + 1)) z = 0

This does not yield the correct formulation. When we recast the original equationas:

(3.11) (1 x)2y 2x(1 x2)y + (l(l + 1)(1 x2)m2) y = 0This easily suggests the following form for our final function:

(3.12) = (1 x2)a(

d

dx

)byn

To find what form a and b take, we consider y1 = x with b = 1 since clearly b mustbe be a positive integer:

= (1 x2)a ddx

(x) = (1 x2)a

= 2ax(1 x2)a1 = 2a(1 x2)a1 + a(a 1)(4x2)(1 x2)a2

(1 x2) 2x +(l(l + 1) m

2

(1 x2))

= 2a(1 x2)a + a(a 1)(4x2)(1 x2)a1 + yx2a(1 x2)a1+l(l+ 1)(1 x2)a m2(1 x2)a1 = 0(3.13)

Upon the expansion of the latter equation for x, we can match each term with itscoefficients. These coefficients must equal zero to satisfy the equation. It is easiestto take the highest coefficients of x:

a(a 1)4x2(1 x2)a1 + 4x2a(1 x2)a1 m2(1 x2)a1 = 0

a2 =m2

4 a = |m|

2(3.14)

Here m=1, but we assume the ansazts of generalization whereby, sans a normaliza-tion constant, the full solution of our angular equation is the associated LegendreFunction,

(3.15) Pml (x) = (1 x2)|m|2

(d

dx

)|m|(1

2ll!

)(d

dx

)l(x2 1)l

3.2. Normalization for final result. Define Akn,s = 11P kn (x)P

ks (x)dx, then by

separation of variables:

Akn,s =

11

(1 x2)k(

dk

dxkPn(x)

)(dk

dxkPs(x)

)dx

=

11

(1 x2)k(

dk

dxkPn(x)

)(dk1

dxk1Ps(x)

)dx

(

dk1

dxk1Ps(x)

)d

dx

((1 x2)k d

k

dxkPn(x)

)dx(3.16)

Introduce z = dk

dxkPn(x). From our previous discussion, we find that,

(3.17)d

dx

((1 x2)z) = ((k(k + 1) l(l + 1)) z = 0

5

Or just as easily,

d

dx

((1 x2)k+1z) = (k(k + 1) l(l+ 1)) (1 x2)kz

= (k l)(k + l+ 1)(1 x2)kz = 0(3.18)

Here we need to compensate for the following: The z used in this previous equationwas built with:

(3.19) z =dk

dxkPn(x) z = d

k1

dxk1Pn(x)

Thus we must shift k k 1. Thus,

Akn,s =

11

P kn (x)Pks (x)dx = (n k + 1)(n+ k)

1

1P k1s Pk1n dx

Akn,s = (n k + 1)(n+ k)Ak1n,sAkn,s = (n k + 1)(n+ k)(n k + 2)(n+ k 1)Ak2n,s =

Akn,s =(n+ k)!

(n k)!A0n,s =

(n+ k)!

(n k)! 11

PnPsdx = ns||Pn||2

We can find ||Pn||2 as another separation of variables chain: 11

Pn(x)dx =

(1

2nn!

) 11

(d

dx

)n(1 x2)n

(d

dx

)m(1 x2)mdx

=

(1

2nn!

)(d

dx

)(1 x2)n

(d

dx

m+1

(1 x2)m|11)

(

1

2nn!

) 11

(d

dx

)m1(1 x2)m

(d

dx

)n+1(1 x2)dx(3.20)

The non-integral part equals zero, the integral component cycles up on the n de-rivative and down on the m derivative. Of course, m = n, but we maintain theseparation for the sake of derivation. The final result of the cycle gives:

(3.21)

11

||Pn(x)||2dx = (2n)!(2nn!)2

11

(d

dx

0

(x2 1)m)dx

To solve the latter integral, we introduce two identities:

(1 x2)m =( x

0

)mPm(x

) (dx)m

1

2n+ 1

(P n+1(x) P n1

)=

1

2n+ 1

1

2nn!

dn

dxn

(1

2(n+ 1)

d2

dx2(x2 1)n+1 2n(x2 1)n1

)

H1

2(n+ 1)

d2

dx2(x2 1)n+1 = (x2 1)n + 2x2n(x2 1)n1

1

2n+ 1

(P n+1(x) P n1(x)

)=

1

2n+ 1

1

2nn!

dn

dxn((x2 1)n + 2n(x2 1)(x2 1)n1)

=1

2nn!

dn

dxn(x2 1) = Pn(x)(3.22)

6

So we follow through,

Pn(x) =1

2n+ 1

(P n+1(x) P n1(x)

) x

0

Pn(x)dx =1

2n+ 1(Pn+1(x) Pn1(x)) x

0

x0

Pn(x)dx dx =1

2n+ 1

(1

2n+ 3(Pn+2(x) Pn) 1

2n 1 (Pn(x) Pn2(x)))

We follow through this derivation cycle n times, splitting each P value as per theabove identity. The Pn(x) integral becomes the (1 x2)n as identified above andhe array Pn will be spread from values of P2n P0. If we integrate both sidesof the equation over the Legendre function range, [-1,1], all said values of Pn willvanish except P0. The prefactors from the expansion give:

2nn!

11

(1 x2)ndx

=

11

2nn!

(( x0

)nPn(x) (dx)

n

)dx =

(2n)!

(2n+ 1)!

11

P0(x)dx(3.23)

But properly, 11

|Pn|2dx = (2n)!(2nn!)2

11

(1 x2)ndx

=(2n)!

(2nn!)2

11

2nn!

( x0

Pn(x) (dx)n

)dx

=(2n)!

(2nn!)2

11

2nn!2nn!

(2n+ 1)!dx

=2

2n+ 1(3.24)

From this we conclude,

(3.25) Akn,s =

((n+ k)!

(n k)!2

2n+ 1

)

Of course, we cannot neglect the symmetric components of theta, which give us anadditional normalization of 2, thus the final normalization condition,

(3.26)

2pi0

pi0

||Y kn | sin()dd

Re-tagging our variables appropriately, the normalized Associated Legendre Poly-nomial is,

Y kn (, ) =

2n+ 1

4

(n k)!(n+ k)!

eim(1 x2) |m|2(

d

dx

)|m|(1

2ll!

)(d

dx

)l(x2 1)l

Or in simplified notation,

(3.27) Y mn (, ) =

2n+ 1

4

(nm)!(n+m)!

eimPml (cos())

.7

4. Radial side

Granted that = l(l + 1), the radial equation becomes simply:

(4.1)d

dr

(r2dR

dr

) 2mr

2

~2(V (r) E)R = l(l+ 1)R

When finding solutions to such equations, it is generally in our best interest toisolate the leading derivatives. We can do so with the substitution:

u(r) = rR(r) R = ur,dR

dr=

r dudr ur2

ddr(r2

dR

dr) = r

d2u

dr2

~2

2m

d2u

dr2+

(V E + ~

2l(l+ 1)

2mr2

)u = 0

H V = e2

4or, =

2mE~

~2

2m

d2u

dr2+

( e

2

4or+

2~2

2m+~2l(l+ 1)

2mr2

)u = 0

1

2d2u

dr2=

(1 me

2

2o~22r+

l(l+ 1)

2r2

)u

H = r , o =me2

2o~2

d2u

d2=

(1 o

+

l(l+ 1)

2

)u(4.2)

Here we have considered the bound states of the electron (of course) where E < 0.Typically, one looks and supposes asymptotic solutions. For ,

(4.3)d2u

d2= u u() = e + e

But of course, finiteness requires that = 0 As 0, the 2 term dominates

(4.4)d2u

d2=

l(l+ 1)u

2

The solution to this equation is interesting to derive. We suppose the existence ofsome function z(), and rewrite the equation thereby:

du

d=

du

dz

dz

d

d2u

d2=

d

d

(d

dzudz

d

)

=dz

d

d

dz

(du

dz

)+

du

dz

d2z

d2

=d2u

dz2

(dz

d

)2+

du

dz

d2z

d2

d2u

d2=

l(l+ 1)u

2

0 =d2u

dz2

(dz

d

)2+

du

dz

d2z

d2 l(l+ 1)u

2(4.5)

8

The trick here is to assume that there exist a z() such that the previous equationis a differential equation with constant coefficients, i.e.:

(dzd

)2l(l+1)2

= 1

dz

d=

l(l + 1)

z =l(l+ 1) ln() = ln(p

l(l+1))(4.6)

Our characteristic equation follows from:

(4.7)l(l + 1)

2u

l(l + 1)

2u l(l+ 1)

2= 0

It is easy to show the resulting characteristic is (1 (2l + 1))/(2l(l + 1)) with

solutions:

u = exp(2(l+ 1)z

2l(l+ 1)

) + iexp(2lz

2l(l+ 1)

)

= exp(2(l + 1)

l(l + 1)ln

2l(l + 1)

) + iexp(2l

l(l+ 1)ln

2l(l+ 1)

)

= exp(ln(l+1)) + iexp(lnl)

= l+1 + il(4.8)

Since this solution contains the zero point, i = 0, and we can write our asymptoticgeneral solution in total as u() = l+1ev(), where v() is some unknown func-tion which properly completes the radial function and is to be found. Reformulatingthe equation thus far,

u() = l+1ev()

du

d= le

((l + 1 )v + dv

dp

)d2u

d2= le

((2l 2 + + l(l+ 1)

)v + 2(l + 1 )dv

d+

d2v

d2

)

With the radial equation as,

(4.9)d2u

d2=

(1 o

+

l(l+ 1)

2

)u

We can plug in the values we found above to reformulate the equation in terms ofv. Via algebra which can be easily done in mind, we find:

(4.10) d2v

d2+ 2(l + 1 )dv

d+ (o 2(l + 1))v = 0

Following typical derivations (though we will part with it eventually), we assumea solution of a series solution form, v() =

j=0 ajp

j . Done so, the prior equation9

becomes,

j(j + 1)aj+1p

j + 2(l + 1)

j=0

aj+1pj 2

j=0

jajpj

+(po 2(l+ 1))j=0

ajpj = 0

j(j + 1)aj+1 + 2(l+ 1)(j + 1)aj+1 2jaj + (po 2(l + 1)) aj = 0 aj+1 =

(2(j + l + 1) o(j + 1)(j + 2l + 2)

)aj

H limj

(2(j + l + 1) o(j + 1)(j + 2l+ 2)

)aj 2

j

j!

() =

j=0

2j

j!pj = e2p @ (4.11)

Obviously the series must terminate dynamically, i.e.

jmax ajmax+1 = 0 2(jmax + l + 1) o = 0 H n = jmax + l + 1 o = 2nOur new recursion equation is:

(4.12) aj+1 =2(j + l+ 1 2n)(j + 1)(j + 2l+ 2)

aj

We consider a few sample results of vln. E.g.

v02 = ao aoxv13 = ao

ao2x

vo3 = ao 2aox+2ao3x2

vl4 = ao aox+2ao10

We can go on ad infinitum, but we seek an analytical solution to this equation. Theasymptotically suggested form gives us a starting point for its completing function.We also suspect that the solution might follow the onion-derivative similar to thatof the angular solution. With these educated guesses, we start as follows: Weassume the solution has a kernel w = exxq

w = exxq + qexxq1w = exxq qexxq1 + q(q 1)exxq2 qexxq1

e.g. @ q = 2 w = 2ex 4exx+ exx2,exw(q = 2) = 2 4x+ x2(4.13)

We had plugged in p = 2 as a simple test of low level results. Take the derivative ofthe latter equation and find, 2x 4 a la v31 = ao ao2 x. One could follow throughthis examination ad infinitum, but we already know the ending to this story so weassume immediately that we have happened upon the correct solution form, i.e.

(4.14) vpqp(x) = k

(d

dx

)p(ex(

d

dx

)qexxq

)

When k = (1)p, the above equation is the associated Laguerre Polynomial, Lpqp.We can bring this all home as follows. We consider our v31 example. In that case,for n = 3, l = 1, we can write the the v completing function with coefficients

10

in terms of the recursion variables. Using v() = L2l+1nl1(2), which, due to the

multiplication by two matches our v31 example,

Now R = u()r , where:

(4.15) u() = l+1eL2l+1nl1(2)

But with o = 2n, and

o =me2

2o~2

(4.16) =me2

4no~2

Let

a =4o~

2

me2

which is the Bohr Radius we had previously derived. Then we write = 1an . Thus = r = ran . Ergo

(4.17) R(r) =N

an

( ran

)lexp( r

na)L2l+1nl1

(2r

na

)

The normalization factor N is determined as follows.

4.1. Normalization of the Radial Term. Key to normalizing the radial term isthe normalization of the Laguerre Polynomials. This task will not be as facile aswas normalization for the Legendre Polynomials.

Given that Ln(x) =ex

n!dn

dx2 (xnex), we note the following:

nLn1(x) =nex

(n 1)!dn1

dxn1xn1ex

=ex

(n 1)!dn1

dxn1(nxn1ex)

=ex

(n 1)!dn1

dxn1(d

dxxnex + xnex)

=nex

n!

dn

dxnxnex +

ex

(n 1)!dn1

dxn1xnex

= nLn(x) +ex

(n 1)!dn1

dxn1xnex(4.18)

In the same way, we find that:

(n+ 1)Ln+1(x) = (n+ 1)ex

(n+ 1)!

dn+1

dxn+1xn+1ex

Hd

dxxn+1ex = (n+ 1)xnex xn+1ex

(n+ 1)Ln+1(x) =ex

n!

dn

dxn((n+ 1)xnex xn+1ex)

=ex

n!

dn

dxn((n+ 1)xnex

) exn!

dn

dxn(xn+1ex)

= (n+ 1)Ln(x) ex

n!

dn

dxn(xn+1ex)(4.19)

11

One final note is to demonstrate the following for one case and assume the rest ofthe cases by assumed induction:

ex

(n 1)!dn1

dxn1(xnex) e

x

n!

dn

dxn(xn+1ex) = ?

n = 1 ex(xex) ex ddx

(x2ex) = ex(xex 2xex + x2ex)

= x2 x = xL1Generally:

ex

(n 1)!dn1

dxn1(xnex) e

x

n!

dn

dxn(xn+1ex) = xLn(x)

Putting the above three results into place, we find that:

(4.20) (n+ 1)Ln+1(x) = (2n+ 1 x)Ln(x) nLn1(x)We can derive another important and useful identity as follows:

xLn(x) = ex

(n 1)!dn1

dxn1(xnex) e

x

n!

dn

dxn(xn+1ex)

xLn(x) Ln = xLn + nLn (n+ 1)Ln+1(x)H xLn(x) = (n+ 1)Ln+1(x) + nLn1(x) (2n+ 1)Ln(x) xLn(x) = nLn(x) nLn1(x)(4.21)

But we are using associated Laguerre Polynomials, so we need to extend theseidentities. We will not work out the details, but taking the derivatives of theprevious identities gives us the associated identities:

(n+ 1)Lkn+1(x) = (2n+ k + 1 x)Lkn(x) (n+ k)Lkn1(x)xd

dxLkn(x) = nL

kn(x) (n+ k)Lkn1(x)(4.22)

The solution we found for the radial equation took the form exp(x/2)x(k+1)/2Lkn(x).The normalization equation for this formulation gives:

0

exxk+1(Lkn(x))2dx =

0

x

n!

dn

dxn(exxn+k

) (Lkn(x)

)dx(4.23)

Applying integration by parts n times, cycling up on the left derivative and downon the right Laguerre, using the derivative formula above, we land at:

(4.24)

0

exxk+1(Lkn(x))2dx =

0

x(n1)

n!(exxn+k)(n+ k)!Lk0(x)dx

(4.25)

0

exxk+1(Lkn(x))2dx =

0

(n+ k)!

n!(exxk)Lk0(x)dx

From the identity, we have:

(4.26) xLk0(x) = (2n+ k + 1)Lk0(x) + (n+ 1)L

k1(x) (n+ k)Lk1(x)

When we plug this into the integral to wash away the x, all terms integrate to zeroexcept Lk0 , and our integral becomes:

(4.27)

0

exxk+1(Lkn(x))2dx = (2n+ k + 1)

(n+ k)!

n!

0

exxkLk0(x)dx

12

With Lk0(x) = k!, and a integration by parts performed on the remaining integral(cycling down the xk so that the k! values cancel, we get:

0

exxk+1(Lkn(x))2dx

= (2n+ k + 1)(n+ k)!

n!

0

exdx =(n+ k)!

n!(2n+ k + 1)(4.28)

Now when we adjust this representation for our own, i.e. n n l 1 andk 2l + 1, the normalization constant becomes (n+l)!(nl1)! (2n). Incorporating thefact that we must convert x 2na , this calls upon an additional normalizationfactor of

(2na

)3/2Thus the radial normalization is:

(4.29) N =

(n l 1)!2n(n+ l)!

(2

na

)3Our final radial equation becomes:

(4.30) R(r) =

(n l 1)!2n(n+ l)!

(2

na

)3e

r

na

(2r

na

)lL2l+1nl1(

2r

na)

5. Final Form

Putting these pieces together, the full function that describes the most basichydrogen quantum model is given as,

(5.1) nlm =

(2

na

)3(n l 1)!2n((n+ 1)!)

e4

na

(2r

na

)lL2l+1nl1

(2r

na

)Y ml (, )

We present this equation in full below.

References

[1] Charles S. Johnson, Lee G. Pedersen, Problems and Solutions in Quantum Chemistry andPhysics, Dover Publications, New York, 1986

[2] Hancs C. Ohanian, Modern Physics, 2nd ed., Prentice Hall 1987, 1995[3] Paul Tipler, Modern Physics, Worth Publishers, 1969, 1978[4] David Griffiths,Introduction to Electrodynamics, 3rd ed., Prentice Hall, 1981, 1989, 1999[5] George B. Arfken, Hans J. Weber, Mathematical Methods For Physicists, 5th ed., Harcourt

Academic Press 2001, 1995, 1985, 1970, 1966[6] David Griffiths, Introduction to Quantum Mechanics, Prentice Hall 1995[7] Arsenin, Basic Equations and Special Functions of Mathematical Physics

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